\(\int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx\) [203]

Optimal result

Integrand size = 23, antiderivative size = 87 \[ \int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\frac {\sqrt {c} \sqrt {a+b x^2} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right ),1-\frac {b c}{a d}\right )}{a \sqrt {d} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}} \]

[Out]

(1/(1+d*x^2/c))^(1/2)*(1+d*x^2/c)^(1/2)*EllipticF(x*d^(1/2)/c^(1/2)/(1+d*x^2/c)^(1/2),(1-b*c/a/d)^(1/2))*c^(1/
2)*(b*x^2+a)^(1/2)/a/d^(1/2)/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(d*x^2+c)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {429} \[ \int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\frac {\sqrt {c} \sqrt {a+b x^2} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right ),1-\frac {b c}{a d}\right )}{a \sqrt {d} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}} \]

[In]

Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]),x]

[Out]

(Sqrt[c]*Sqrt[a + b*x^2]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(a*Sqrt[d]*Sqrt[(c*(a + b*x^
2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2])

Rule 429

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]/(a*Rt[d/c, 2]*
Sqrt[c + d*x^2]*Sqrt[c*((a + b*x^2)/(a*(c + d*x^2)))]))*EllipticF[ArcTan[Rt[d/c, 2]*x], 1 - b*(c/(a*d))], x] /
; FreeQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {c} \sqrt {a+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{a \sqrt {d} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.99 \[ \int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\frac {\sqrt {\frac {a+b x^2}{a}} \sqrt {\frac {c+d x^2}{c}} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {-\frac {b}{a}} x\right ),\frac {a d}{b c}\right )}{\sqrt {-\frac {b}{a}} \sqrt {a+b x^2} \sqrt {c+d x^2}} \]

[In]

Integrate[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]),x]

[Out]

(Sqrt[(a + b*x^2)/a]*Sqrt[(c + d*x^2)/c]*EllipticF[ArcSin[Sqrt[-(b/a)]*x], (a*d)/(b*c)])/(Sqrt[-(b/a)]*Sqrt[a
+ b*x^2]*Sqrt[c + d*x^2])

Maple [A] (verified)

Time = 3.01 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.15

method	result	size
default	\(\frac {F\left (x \sqrt {-\frac {b}{a}}, \sqrt {\frac {a d}{b c}}\right ) \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}}{\sqrt {-\frac {b}{a}}\, \left (b d \,x^{4}+a d \,x^{2}+c b \,x^{2}+a c \right )}\)	\(100\)
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, F\left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )}{\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\, \sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+c b \,x^{2}+a c}}\)	\(122\)

[In]

int(1/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

EllipticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*((d*x^2+c)/c)^(1/2)*((b*x^2+a)/a)^(1/2)*(d*x^2+c)^(1/2)*(b*x^2+a)^(1
/2)/(-b/a)^(1/2)/(b*d*x^4+a*d*x^2+b*c*x^2+a*c)

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.48 \[ \int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=-\frac {\sqrt {a c} \sqrt {-\frac {b}{a}} F(\arcsin \left (x \sqrt {-\frac {b}{a}}\right )\,|\,\frac {a d}{b c})}{b c} \]

[In]

integrate(1/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

-sqrt(a*c)*sqrt(-b/a)*elliptic_f(arcsin(x*sqrt(-b/a)), a*d/(b*c))/(b*c)

Sympy [F]

\[ \int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\int \frac {1}{\sqrt {a + b x^{2}} \sqrt {c + d x^{2}}}\, dx \]

[In]

integrate(1/(b*x**2+a)**(1/2)/(d*x**2+c)**(1/2),x)

[Out]

Integral(1/(sqrt(a + b*x**2)*sqrt(c + d*x**2)), x)

Maxima [F]

\[ \int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\int { \frac {1}{\sqrt {b x^{2} + a} \sqrt {d x^{2} + c}} \,d x } \]

[In]

integrate(1/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(b*x^2 + a)*sqrt(d*x^2 + c)), x)

Giac [F]

\[ \int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\int { \frac {1}{\sqrt {b x^{2} + a} \sqrt {d x^{2} + c}} \,d x } \]

[In]

integrate(1/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(b*x^2 + a)*sqrt(d*x^2 + c)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\int \frac {1}{\sqrt {b\,x^2+a}\,\sqrt {d\,x^2+c}} \,d x \]

[In]

int(1/((a + b*x^2)^(1/2)*(c + d*x^2)^(1/2)),x)

[Out]

int(1/((a + b*x^2)^(1/2)*(c + d*x^2)^(1/2)), x)